CHAPTER 3:KINETICS OF PARTICLES

SAMPLE PROBLEM 3/1,PAGE NUMBER:126

In [1]:
import math

# Variable Declaration
m_m=75;# The mass of the man in kg
m_t=750;# The total mass of the elevator in kg
T=8300;# The tension T in the hoisting cable in N
t_ll=0;# Lower limit of integration
t_ul=3;# Upper limit of integration
g=9.81;# The acceleration due to gravity in m/s**2

# Calculation
W=m_t*g;# The total weight of the elevator
# From the free-body diagram of the elevator,SigmaF_y=m*a_y
SigmaF_y=T-W;# N
a_y=SigmaF_y/(m_t);# The acceleration in m/s**2
# From the free-body diagram of the man,SigmaF_y=m*a_y
W_m=m_m*g;# Weight of the man in N
R=(m_m*a_y)+W_m;# The equal and opposite reaction R in N
# The velocity reached at the end of the 3 seconds is delv=integral(a.dt),integrating this equation we get
v=a_y*(t_ul-t_ll);# The velocity in m/s
print"The equal and opposite reaction,R=",R,"N"
print"The velocity reached at the end of the 3 seconds,v=",round(v,2),"m/s"

The equal and opposite reaction,R= 830.0 N
The velocity reached at the end of the 3 seconds,v= 3.77 m/s

SAMPLE PROBLEM 3/2,PAGE NUMBER:126

In [2]:
import math

# Variable Declaration
m=200;# The mass of the car in kg
T=2.4;# The cable tension in kN
g=9.81;# The acceleration due to gravity in m/s**2

# Calculation
W=(m*g)/1000;# The weight of the car in kN
# In y direction,SigmaF_y=0
x=5;#  adjacent side
y=12;#  opposite side
r=13;#   hypotenuse
P=((T*x)/r)+((W*y)/r);# The total force exerted by the supporting cable on the wheels in N
print"The total force exerted by the supporting cable on the wheels,P=",round(P,2),"kN"
# In x direction,SigmaF_x=m*a_x
x=12;# #  adjacent side
y=5;# opposite side
r=13;# #  hypotenuse
a=((((T*1000*x)/r)-((W*1000*y)/r))/m);# #  The acceleration of the car in m/s**2
print"The acceleration of the car,a=",round(a,2),"m/s**2"

The total force exerted by the supporting cable on the wheels,P= 2.73 kN
The acceleration of the car,a= 7.3 m/s**2

SAMPLE PROBLEM 3/3,PAGE NUMBER:127

In [3]:
import math
from scipy.optimize import fsolve
# Variable declaration
m_A=250;# The mass of concrete block A in lb
m=400;# lb
theta=30;# degree
mu_k=0.5;# The coefficient of kinetic friction between the log and the ramp
x=20;# ft
g=32.2;# The acceleration due to gravity in ft/sec**2

# Calculation
# SigmaF_y=0;
N=m*math.cos(theta*math.pi/180);# lb
# SigmaF_x=ma_x;
def equations(p):
    T,a_A,a_C=p
    return((0-((2*a_C)+a_A)),(((mu_k*N)-(2*T)+(m*math.sin(theta*math.pi/180)))-((m/g)*a_C)),((m_A-T)-((m_A/g)*a_A)))
T,a_A,a_C=fsolve(equations,(100,1,1))
v_A=math.sqrt(2*a_A*x);# ft/sec
print"\nThe velocity of the block as it hits the ground at B,v_A=%2.2f ft/sec"%v_A;

The velocity of the block as it hits the ground at B,v_A=15.27 ft/sec

SAMPLE PROBLEM 3/4,PAGE NUMBER:128

In [4]:
import math
import sympy

# Variable Declaration
m=10;# The mass in kg
v=2;# The speed in m/s
R=8;# N

# Calculation
t = sympy.Symbol('t')
k=R/v**2;# N.s**2/m**2
# SigmaF_x=ma_x;
v_0=v;# m/s
v=v_0/2;# m/s
t_r=5*((1/v)-(0.5));# The time requirwd to reduce the speed in s
print"\nThe time t required for it to reduce its speed to 1 m/s,t=%1.2f s"%t_r
t_0=0;# s
t_1=2.5;# s
x=sympy.integrate((10/(5+(2*t))),('t',t_0,t_1));
print"\nThe corresponding travel distance,x=%1.2f m"%x;

The time t required for it to reduce its speed to 1 m/s,t=2.50 s

The corresponding travel distance,x=3.47 m

SAMPLE PROBLEM 3/8,PAGE NUMBER:140

In [5]:
import math
import numpy

# Variable Declaration
m=1500;# The mass of the car in kg
v_A=100;# The velocity in km/h
v_C=50;# The velocity in km/h
rho_A=400;# The radius of curvature in m
rho_C=80;# The radius of curvature in m
delta_s=200;# m

# Calculation
a_t=abs((((v_C/3.6)**2)-((v_A/3.6)**2))/(2*delta_s));# The tangential acceleration in m/s**2
a_na=((v_A/3.6)**2)/rho_A;# The normal components of acceleration at A in m/s**2
a_nb=0;# The normal components of acceleration at B in m/s**2
a_nc=((v_C/3.6)**2)/rho_C;# The normal components of acceleration at C in m/s**2
F_t=m*a_t;# N
F_na=m*a_na;# N
F_nb=m*a_nb;# N
F_nc=m*a_nc;# N
F_a=math.sqrt(F_na**2+F_t**2);# The total horizontal force acting on the tires at A in N
F_b=math.sqrt(F_nb**2+F_t**2);# The total horizontal force acting on the tires at B in N
F_c=math.sqrt(F_nc**2+F_t**2);# The total horizontal force acting on the tires at C in N
print"\nAt A,F=%4.0f N \nAt B,F=%4.0f N \nAt C,F=%4.0f N"%(F_a,F_b,F_c);

At A,F=3617 N 
At B,F=2170 N 
At C,F=4218 N

SAMPLE PROBLEM 3/9,PAGE NUMBER:141

In [6]:
import math
# Variable Declaration
h=200;# The altitude in mi
R=3959;# mi
g=32.234;# The acceleration due to gravity in ft/sec**2

# Calculation
# SigmaF_n=ma_n;
v=(R*5280)*math.sqrt(g/((R+h)*5280));# ft/sec
print"\nThe velocity required for the spacecraft,v=%5.0f ft/sec"%v;

The velocity required for the spacecraft,v=25326 ft/sec

SAMPLE PROBLEM 3/11,PAGE NUMBER:162

In [7]:
import math
# Variable Declaration
m=50;# kg
v_1=4;# m/s
mu_k=0.30;# The coefficient of kinetic friction 
g=9.81;# The acceleration due to gravity in m/sec**2
s=10;# m
theta=15;# degree
R=474;# N

# Calculation
U_12=((m*g)*s*math.sin(theta*math.pi/180))-(mu_k*R*(s));# The total work done on the crate during the motion in J
v_2=math.sqrt((((m*v_1**2)/2)+U_12)/((m/2)));# The velocity of the crate in m/s
print"\nThe velocity of the crate,v_2=%1.2f m/s"%v_2;

The velocity of the crate,v_2=3.15 m/s

SAMPLE PROBLEM 3/12,PAGE NUMBER:162

In [8]:
import math
# Variable Declaration
m=80;# kg
v=72;# km/h
s=75;# m
mu_sa=0.30;# The coefficient of static friction 
mu_ka=0.28;# The coefficient of kinetic friction
mu_sb=0.25;# The coefficient of static friction 
mu_kb=0.20;# The coefficient of kinetic friction
g=9.81;# The acceleration due to gravity in m/sec**2

# Calculation
a=(v/3.6)**2/(2*s);# m/s**2
# Case(a)
F=m*a;# The friction force on the block in N
U_12=F*s;# The work done in J
print"\n(a)The work done by the friction force acting on the crate,U_12=%5.0f J (or) %2.0f kJ"%(U_12,U_12/1000);
# Case(b)
F=mu_sb*m*g;# N
F=mu_kb*m*g;# N
a_b=F/m;# The acceleration in m/s**2
deltaS=(a_b/a)*s;# The displacement of a crate in m
U_12=F*deltaS;# The work done in J
print"\n(b)The work done by the friction force acting on the crate,U_12=%4.0f J (or) %1.2f kJ"%(U_12,U_12/1000);

(a)The work done by the friction force acting on the crate,U_12=16000 J (or) 16 kJ

(b)The work done by the friction force acting on the crate,U_12=8661 J (or) 8.66 kJ

SAMPLE PROBLEM 3/13,PAGE NUMBER:163

In [9]:
import math
# Variable Declaration
m=50;# The mass of the block in kg
F=300;# N
x_1=0.233;# m
k=80;# The spring stifness in N/m
x=1.2;# m
y=0.9;# m

# Calculation
x_2=x_1+x;# m
U_12=(1/2)*k*(x_1**2-x_2**2);# The work done by the spring force acting on the block in J
s=math.sqrt(x**2+y**2)-y;# m
W=F*s;# The work done in J
T_1=0;# J
v=math.sqrt(((U_12+W)*2)/m);# m/s
print"\nThe velocity of the block as it reaches position B,v=%1.2f m/s"%v;

The velocity of the block as it reaches position B,v=2.68 m/s

SAMPLE PROBLEM 3/14,PAGE NUMBER:164

In [10]:
import math
# Variable Declaration
F=800;# lb
theta=30;# degree
v=4;# ft/sec
P=6;# hp
P_i=8;# hp
g=32.2;# The acceleration due to gravity in ft/sec**2

# Calculation
N=(F*math.sin(theta*math.pi/180));# lb
# SigmaF_x=0;
T=(P*550)/v;# The tension in the cable in N
mu_k=(T-(F*math.sin(theta*math.pi/180)))/N;# The coefficient of kinetic friction
T=(P_i*550)/v;# lb
a=(g*(T-(N*mu_k)-(F*math.sin(theta*math.pi/180))))/F;# The acceleration in ft/sec**2
print"\nThe corresponding instantaneous acceleration of the log,a=%2.2f ft/sec**2"%a;

The corresponding instantaneous acceleration of the log,a=11.07 ft/sec**2

SAMPLE PROBLEM 3/15,PAGE NUMBER:180

In [11]:
import math
# Variable Declaration
h_1=500;# km
v_1=30000;# km/h
h_2=1200;# km
g=9.81;# The acceleration due to gravity in m/sec**2
R=6371;# km

# Calculation
v_2=math.sqrt((v_1/3.6)**2+(2*g*(R*10**3)**2*((10**-3/(R+h_2))-(10**-3/(R+h_1)))));
print"\nThe velocity of the satellite as it reaches point B,v_2=%4.0f m/s (or) v_2=%5.0f km/h"%(v_2,v_2*3.6);

The velocity of the satellite as it reaches point B,v_2=7663 m/s (or) v_2=27588 km/h

SAMPLE PROBLEM 3/16,PAGE NUMBER:180

In [12]:
import math
# Variable Declaration
mg=6;# lb
k=2;# lb/in
g=32.2;# The acceleration due to gravity in ft/sec**2
h=24;# in
x_1=24/12;# ft
x_2=(((24*math.sqrt(2))/12)-(24/12));# ft

# Calculation
# The reaction of the rod on the slider is normal to the motion and does no work.
T_1=0;# ft-lb
U_12=0;# ft-lb
# We define the datum to be at the level of position 1, so that the gravitational potential energies are
V_1g=0;# ft-lb
V_2g=-(mg)*(h/12);# ft-lb
V_1e=((k*12)*(x_1)**2)/2;# ft-lb
V_2e=((k*12)*(x_2)**2)/2;# ft-lb
v_2=math.sqrt(((T_1+(V_1g+V_1e)+U_12)-(V_2g+V_2e))*(2*(g/mg)));# ft/sec
print"\nThe velocity of the slider as it passes position 2,v_2=%2.1f ft/sec"%v_2;

The velocity of the slider as it passes position 2,v_2=23.6 ft/sec

SAMPLE PROBLEM 3/17,PAGE NUMBER:180

In [13]:
import math
# Variable Declaration
m=10;# kg
k=60;# N/m
F=250;# N
theta=30;# degree
ABbar=1.5;# m
BCbar=0.9;# m
g=9.81;# The acceleration due to gravity in m/sec**2
d_AC=1.2;# The distance in m
d_BC=0.9;# The distance in m

# Calculation
s=ABbar-BCbar;# m
U_ac=F*s;# J
V_Ag=0;# The initial gravitational potential energy in J
T_A=(1/2)*m*V_Ag**2;# N.m
V_Cg=m*g*(d_AC*math.sin(theta*math.pi/180));# The final gravitational potential energy in J
x_A=s;# m
x_B=s+d_AC;# m
V_Ae=(k*(x_A)**2)/2;# The initial elastic potential energy in J
V_Ce=(k*(x_B)**2)/2;# The final elastic potential energy in J
# Substitution into the alternative work-energy equation 3/21a gives
v_c=math.sqrt((((T_A+V_Ag+V_Ae+U_ac)-(V_Cg+V_Ce))*2)/m);# m/s
print"\nThe velocity of the slider as it passes point C,v_C=%0.3f m/s"%v_c;

The velocity of the slider as it passes point C,v_C=0.974 m/s

SAMPLE PROBLEM 3/18,PAGE NUMBER:181

In [14]:
%matplotlib inline
import math
from scipy.optimize import fsolve
import matplotlib.pyplot as plt
from matplotlib.pyplot import plot,suptitle,xlabel,ylabel
# Variable Declaration
m_A=2;# kg
m_B=4;# kg
L=0.5;# m
K_theta=13;# N.m/rad
g=9.81;# The acceleration due to gravity in m/sec**2

# Calculation
# (a)
# T_1+V_1+U_12=T_2+V_2
def eqn(p):
    v_A=p
    return((((1*2**-1)*m_A*v_A**2)+((1*2**-1)*m_B*(v_A/4)**2)-(m_A*g*L)-(m_B*g*(L*math.sqrt(2)*4**-1))+((1*2**-1)*K_theta*(math.pi/2)**2))-0);
v_A=fsolve(eqn,(1));# m/s
print"\nThe speed of particle A,v_A=%0.3f m/s"%v_A;
# (b)
theta=[0,10,20,30,40,50,60,70,80,90];
v_A1=(math.sqrt(((m_A*g*L*(1-math.cos(theta[0]*math.pi*180**-1)))+((m_B*g*(1*2**-1)*(((L*math.sqrt(2))*2**-1)-((2*(L*2**-1)*math.sin((((90-(theta[0]))*2**-1))*math.pi*180**-1))))))-((1*2**-1)*K_theta*(theta[0]*(math.pi*180**-1))**2))/(((1*2**-1)*m_A)+((1*2**-1)*m_B*((1*4**-1)*math.cos((((90-(theta[0])))*2**-1)*math.pi*180**-1))**2))))
v_A2=(math.sqrt(((m_A*g*L*(1-math.cos(theta[1]*math.pi*180**-1)))+((m_B*g*(1*2**-1)*(((L*math.sqrt(2))*2**-1)-((2*(L*2**-1)*math.sin((((90-(theta[1]))*2**-1))*math.pi*180**-1))))))-((1*2**-1)*K_theta*(theta[1]*(math.pi*180**-1))**2))/(((1*2**-1)*m_A)+((1*2**-1)*m_B*((1*4**-1)*math.cos((((90-(theta[1])))*2**-1)*math.pi*180**-1))**2))))
v_A3=(math.sqrt(((m_A*g*L*(1-math.cos(theta[2]*math.pi*180**-1)))+((m_B*g*(1*2**-1)*(((L*math.sqrt(2))*2**-1)-((2*(L*2**-1)*math.sin((((90-(theta[2]))*2**-1))*math.pi*180**-1))))))-((1*2**-1)*K_theta*(theta[2]*(math.pi*180**-1))**2))/(((1*2**-1)*m_A)+((1*2**-1)*m_B*((1*4**-1)*math.cos((((90-(theta[2])))*2**-1)*math.pi*180**-1))**2))))
v_A4=(math.sqrt(((m_A*g*L*(1-math.cos(theta[3]*math.pi*180**-1)))+((m_B*g*(1*2**-1)*(((L*math.sqrt(2))*2**-1)-((2*(L*2**-1)*math.sin((((90-(theta[3]))*2**-1))*math.pi*180**-1))))))-((1*2**-1)*K_theta*(theta[3]*(math.pi*180**-1))**2))/(((1*2**-1)*m_A)+((1*2**-1)*m_B*((1*4**-1)*math.cos((((90-(theta[3])))*2**-1)*math.pi*180**-1))**2))))
v_A5=(math.sqrt(((m_A*g*L*(1-math.cos(theta[4]*math.pi*180**-1)))+((m_B*g*(1*2**-1)*(((L*math.sqrt(2))*2**-1)-((2*(L*2**-1)*math.sin((((90-(theta[4]))*2**-1))*math.pi*180**-1))))))-((1*2**-1)*K_theta*(theta[4]*(math.pi*180**-1))**2))/(((1*2**-1)*m_A)+((1*2**-1)*m_B*((1*4**-1)*math.cos((((90-(theta[4])))*2**-1)*math.pi*180**-1))**2))))
v_A6=(math.sqrt(((m_A*g*L*(1-math.cos(theta[5]*math.pi*180**-1)))+((m_B*g*(1*2**-1)*(((L*math.sqrt(2))*2**-1)-((2*(L*2**-1)*math.sin((((90-(theta[5]))*2**-1))*math.pi*180**-1))))))-((1*2**-1)*K_theta*(theta[5]*(math.pi*180**-1))**2))/(((1*2**-1)*m_A)+((1*2**-1)*m_B*((1*4**-1)*math.cos((((90-(theta[5])))*2**-1)*math.pi*180**-1))**2))))
v_A7=(math.sqrt(((m_A*g*L*(1-math.cos(theta[6]*math.pi*180**-1)))+((m_B*g*(1*2**-1)*(((L*math.sqrt(2))*2**-1)-((2*(L*2**-1)*math.sin((((90-(theta[6]))*2**-1))*math.pi*180**-1))))))-((1*2**-1)*K_theta*(theta[6]*(math.pi*180**-1))**2))/(((1*2**-1)*m_A)+((1*2**-1)*m_B*((1*4**-1)*math.cos((((90-(theta[6])))*2**-1)*math.pi*180**-1))**2))))
v_A8=(math.sqrt(((m_A*g*L*(1-math.cos(theta[7]*math.pi*180**-1)))+((m_B*g*(1*2**-1)*(((L*math.sqrt(2))*2**-1)-((2*(L*2**-1)*math.sin((((90-(theta[7]))*2**-1))*math.pi*180**-1))))))-((1*2**-1)*K_theta*(theta[7]*(math.pi*180**-1))**2))/(((1*2**-1)*m_A)+((1*2**-1)*m_B*((1*4**-1)*math.cos((((90-(theta[7])))*2**-1)*math.pi*180**-1))**2))))
v_A9=(math.sqrt(((m_A*g*L*(1-math.cos(theta[8]*math.pi*180**-1)))+((m_B*g*(1*2**-1)*(((L*math.sqrt(2))*2**-1)-((2*(L*2**-1)*math.sin((((90-(theta[8]))*2**-1))*math.pi*180**-1))))))-((1*2**-1)*K_theta*(theta[8]*(math.pi*180**-1))**2))/(((1*2**-1)*m_A)+((1*2**-1)*m_B*((1*4**-1)*math.cos((((90-(theta[8])))*2**-1)*math.pi*180**-1))**2))))
v_A10=(math.sqrt(((m_A*g*L*(1-math.cos(theta[9]*math.pi*180**-1)))+((m_B*g*(1*2**-1)*(((L*math.sqrt(2))*2**-1)-((2*(L*2**-1)*math.sin((((90-(theta[9]))*2**-1))*math.pi*180**-1))))))-((1*2**-1)*K_theta*(theta[9]*(math.pi*180**-1))**2))/(((1*2**-1)*m_A)+((1*2**-1)*m_B*((1*4**-1)*math.cos((((90-(theta[9])))*2**-1)*math.pi*180**-1))**2))))
v_A=[v_A1,v_A2,v_A3,v_A4,v_A5,v_A6,v_A7,v_A8,v_A9,v_A10];# m/s
plt.plot(theta,v_A,color='red');
xlabel('theta,deg');
ylabel('v_A,m/s');
print"\nThe maximum value of v_A is seen to be (v_A)_max=1.400 m/s at theta=56.4 degree.";
plt.show()

The speed of particle A,v_A=0.794 m/s

The maximum value of v_A is seen to be (v_A)_max=1.400 m/s at theta=56.4 degree.

SAMPLE PROBLEM 3/19,PAGE NUMBER:194

In [15]:
import math
# Variable Declaration
v_1=50;# ft/sec
v_2=70;# ft/sec
theta=15;# degree
dt=0.02;# sec
g=32.2;# The acceleration due to gravity in ft/sec**2

# Calculation
W=2*16**-1;# N
v_1x=v_1;# ft/sec
v_2x=v_2;# ft/sec
v_1y=0;# ft/sec
v_2y=v_2;# ft/sec
R_x=(((W/g)*(v_2x*math.cos(theta*math.pi/180)))+((W/g)*(v_1x)))/dt;# lb
R_y=(((W/g)*(v_2y*math.sin(theta*math.pi/180)))+((W/g)*(v_1y)))/dt;# lb
R=math.sqrt(R_x**2+R_y**2);# lb
beta=math.degrees(math.atan(R_y/R_x));# degree
print"\nThe magnitude of the average force exerted by the racket on the ball,R=%2.1f lb \nThe angle made by R with the horizontal,beta=%1.2f degree"%(R,beta);

The magnitude of the average force exerted by the racket on the ball,R=23.1 lb 
The angle made by R with the horizontal,beta=8.76 degree

SAMPLE PROBLEM 3/20,PAGE NUMBER:196

In [16]:
import math
# Variable Declaration
# G=(3/2)*(t**2+3)j-((2/3)*(t**3-4))k
z=2;# lb
t=2;# sec

# Calculation
F=[(3*t),(2-(2*t**2))];# lb
F_r=math.sqrt(F[0]**2+F[1]**2);# lb
print"\nF=%1.0fj+(%1.0fk)lb \nF=%1.3f lb"%(F[0],F[1],F_r);

F=6j+(-6k)lb 
F=8.485 lb

SAMPLE PROBLEM 3/21,PAGE NUMBER:196

In [17]:
import math
from numpy.linalg import norm
# Variable Declaration
m=0.5;# kg
v_1x=10;# m/s
v_1y=0;# m/s
t_1=1;# s
t_2=2;# s
t_3=3;# s

# Calculation
v_2x=((m*v_1x)-((4*(t_1))+(2*(t_3-t_1))))/(m);# m/s
v_2y=((m*v_1y)+((1*(t_2))+(2*(t_3-t_2))))/(m);# m/s
v_2=[v_2x,v_2y];# m/s
v_2=norm(v_2);# m/s
theta_x=180+math.atan(v_2y/v_2x)*math.pi/180;# degree
print"\nThe velocity of the particle at the end of the 3-s interval,v_2=%2.0f m/s \ntheta_x=%3.1f degree"%(v_2,theta_x);

The velocity of the particle at the end of the 3-s interval,v_2=10 m/s 
theta_x=180.0 degree

SAMPLE PROBLEM 3/22,PAGE NUMBER:196

In [18]:
import math
# Variable Declaration
m=150;#  kg
v_1=4;#  m/s
t_0=0;#  s
t_1=4;#  s
P=600;#  N
t_2=8;#  s
theta=30;#  degree
g=9.81;#  The acceleration due to gravity in m/sec**2

# Calculation
deltat=(m*0)+((m*v_1)-((v_1*2*P)/2)+(m*g*math.sin(theta*math.pi/180)))/((2*P)+(m*g*math.sin(theta*math.pi/180)));#  s
t_a=v_1+deltat;#  s
v_2x=((m*-v_1)+((v_1*2*P)/2)+(v_1*2*P)-(m*g*math.sin(theta*math.pi/180)*t_2))/m;#  m/s
print"\n(a)The time at which the skip reverses its direction,t_a=%1.2f s \n(b)The velocity of the skip,v_2x=%1.2f m/s"%(t_a,v_2x);

(a)The time at which the skip reverses its direction,t_a=3.45 s 
(b)The velocity of the skip,v_2x=4.76 m/s

SAMPLE PROBLEM 3/23,PAGE NUMBER:196

In [19]:
import math
# Variable Declaration
m_1=0.050;#   kg
m_2=4;#   kg
v_1=600;#  m/s
v_2=12;#  m/s
theta=30;#   degree


# Calculation
v_2=[(m_2*v_2*math.cos(theta*math.pi/180))/(m_1+m_2),((m_1*v_1)+(m_2*v_2*math.sin(theta*math.pi/180)))/(m_1+m_2)];#  m/s
v_x=v_2[0];#  m/s
v_y=v_2[1];#  m/s
V_2=math.sqrt((v_x**2+v_y**2));#  m/s
theta=math.atan((v_y/v_x))*math.pi/180;#  degree
print"\nThe velocity of the block and embedded bullet immediately after impact,v_2=%2.2fi+%2.2fj m/s \nThe final velocity and its direction are given by v_2=%2.2f m/s and theta=%2.1f degree"%(v_x,v_y,V_2,theta);

The velocity of the block and embedded bullet immediately after impact,v_2=10.26i+13.33j m/s 
The final velocity and its direction are given by v_2=16.83 m/s and theta=0.0 degree

SAMPLE PROBLEM 3/24,PAGE NUMBER:209

In [20]:
import math
from numpy.linalg import det

# Variable Declaration
F=10;#   N
m=2;#  kg
v=5;#  m/s 
x=3;#  m
y=6;#  m
z=4;#  m

# Calculation
r=[x,y,z];#  m
mv=[0,(m*v),0];#  (kg.m/s)
H_o1=det([[r[1],r[2]],[mv[1],mv[2]]]);#   N.m/s
H_o2=-det([[r[0],r[2]],[mv[0],mv[2]]])#  N.m/s
H_o3=det([[r[0],r[1]],[mv[0],mv[1]]]);#  N.m/s
H_O=[H_o1,H_o2,H_o3];#  N.m/s
F=[0,0,F];#  N.m/s
Hdot_o1=det([[r[1],r[2]],[F[1],F[2]]]);#  N.m
Hdot_o2=-det([[r[0],r[2]],[F[0],F[2]]]);#  N.m
Hdot_o3=det([[r[0],r[1]],[F[0],F[1]]]);#  N.m
Hdot_O=[Hdot_o1,Hdot_o2,Hdot_o3];#  N.m
print"\nThe angular momentum H_O about point O,H_O=%2.0fi+(%2.0f)j+%2.0fk N.m/s \nThe time derivative,Hdot=%2.0fi+(%2.0f)j+%2.0fk N.m"%(H_O[0],H_O[1],H_O[2],Hdot_O[0],Hdot_O[1],Hdot_O[2]);

The angular momentum H_O about point O,H_O=-40i+(-0)j+30k N.m/s 
The time derivative,Hdot=60i+(-30)j+ 0k N.m

SAMPLE PROBLEM 3/25,PAGE NUMBER:209

In [21]:
import math
v_A=740;#  m/s
r_A=6000*10**6;#  km
r_B=75*10**6;#  km

# Calculation
v_B=(r_A*v_A)/r_B;#  m/s
print"\nThe speed of comet at the point B of closest approach to the sun,v_B=%5.0f m/s"%v_B;

The speed of comet at the point B of closest approach to the sun,v_B=59200 m/s

SAMPLE PROBLEM 3/28,PAGE NUMBER:221

In [22]:
import math
# Variable Declaration
m=800;#  kg
g=9.81;#  m/s**2
h=2;#  m
m_p=2400;#  kg
h_1=0.1;#  m

# Calculation
v_r=math.sqrt(2*g*h);#  m/s
v_ra=math.sqrt(2*g*h_1);#  m/s
# (a)
v_pa=(((m*v_r)+0)+(m*v_ra))/m_p;#  m/s
# (b)
e=(v_pa+v_ra)/(v_r+0);#  The coefficient of restitution
# (c)
T=m*g*h;#  J
T_a=((m*v_ra**2)/2)+((m_p*v_pa**2)/2);#  J
E_l=((T-T_a)/T)*100;#  The percentage loss of energy(%)
print"\n(a)The velocity of the pile immediately after impact,v_p=%1.2f m/s \n(b)The coefficient of restitution,e=%0.3f \n(c)The percentage loss of energy due to the impact is %2.1f percentage."%(v_pa,e,E_l);

(a)The velocity of the pile immediately after impact,v_p=2.55 m/s 
(b)The coefficient of restitution,e=0.631 
(c)The percentage loss of energy due to the impact is 45.1 percentage.

SAMPLE PROBLEM 3/29,PAGE NUMBER:221

In [23]:
import math
# Variable Declaration
v_1=50;#   m/s
v_2=0;#   m/s
e=0.5;#  The effective coefficient of restitution
theta=30;#  degree

# Calculation
v_1an=e*v_1*math.sin(theta*math.pi/180);#   ft/sec
v_1at=v_1*math.cos(theta*math.pi/180);#   ft/sec
#  Assume ' as a
v_a=math.sqrt((v_1an)**2+(v_1at)**2);#   ft/sec
theta_a=math.atan((v_1an/v_1at))*math.pi/180;#  degree
print"\nThe rebound velocity and its angle are then v_a=%2.1f ft/sec and theta_a=%2.1f degree"%(v_a,theta_a);

The rebound velocity and its angle are then v_a=45.1 ft/sec and theta_a=0.0 degree

SAMPLE PROBLEM 3/30,PAGE NUMBER:222

In [24]:
import math
from numpy.linalg import solve
# Variable Declaration
v_1=6;#   m/s
v_2=0;#   m/s
e=0.6;#  The coefficient-of-restitution
theta=30;#  degree

# Calculation
v_1n=v_1*math.cos(theta*math.pi/180);#  m/s
v_1t=v_1*math.sin(theta*math.pi/180);#  m/s
v_2n=0;#  m/s
v_2t=v_2n;#  m/s
A=([1,1],[1,-1]);# 
B=([v_1n+0,(e*v_1n)]);# 
v=solve(A,B);# 
v_1an=v[0];#  m/s
v_2an=v[1];#  m/s
v_1at=v_1t;#  m/s
v_2at=v_2t;#  m/s
v_1a=math.sqrt((v_1an)**2+(v_1at)**2);#  m/s
v_2a=math.sqrt((v_2an)**2+(v_2at)**2);#  m/s
thetaa=math.atan((v_1an/v_1at)*math.pi/180)*math.pi/180;#  m/s
# The kinetic energies just before and just after impact, with m=m1=m2,are
T=18;#  m
T_a=13.68;#  m
E_l=((T-T_a)/T)*100;#  The percentage energy loss(%)
print"\nThe final speeds of the particles v_1a=%1.2f m/s ,v_2a=%1.2f m/s \nThe angle which v_1a makes with the t-direction,theta=%2.2f degree \nThe percentage energy loss is %2.0f percentage."%(v_1a,v_2a,thetaa,E_l);

The final speeds of the particles v_1a=5.13 m/s ,v_2a=1.04 m/s 
The angle which v_1a makes with the t-direction,theta=0.00 degree 
The percentage energy loss is 24 percentage.

SAMPLE PROBLEM 3/31,PAGE NUMBER:236

In [25]:
import math
# Variable Declaration
h_1=2000;#   The perigee altitude in km
h_2=4000;#   The apogee altitude in km
h_c=2500;#   The  altitude of the satellite in km
g=9.825;#    The acceleration due to gravity in m/sec**2
R=12742/2;#  km

# Calculation
# (a)
r_max=R+h_2;#  km
r_min=R+h_1;#  km
a=(r_min+r_max)/2;#  km
v_P=(R*10**3*math.sqrt(g*(a*10**3)**-1)*math.sqrt(r_max/r_min));#  m/s
v_A=(R*10**3*math.sqrt(g*(a*10**3)**-1)*math.sqrt(r_min*r_max**-1));#  m/s
# (b)
r=R+h_c;#  km
v_C=math.sqrt((2*g*(R*10**3)**2)*((1*r**-1)-(1*(2*a)**-1))*(1*(10**3)**-1));#  m/s
# (c)
tau=(2*math.pi*((a*10**3)**(3*2**-1)))/((R*10**3)*math.sqrt(g));#  km
tau_h=tau*3600**-1;#  km
print"\n(a)The necessary perigee velocity,v_P=%4.0f m/s (or) %5.0f km/h \n   The necessary apogee velocity,v_A=%4.0f m/s (or) %5.0f km/h \n(b)The velocity at point C,v_C=%4.0f m/s (or) %5.0f km/h \n(c)The period of the orbit,tau=%1.3f h"%(v_P,v_P*3.6,v_A,v_A*3.6,v_C,v_C*3.6,tau_h);

(a)The necessary perigee velocity,v_P=6524 m/s (or) 23485 km/h 
   The necessary apogee velocity,v_A=5861 m/s (or) 21099 km/h 
(b)The velocity at point C,v_C=6881 m/s (or) 24773 km/h 
(c)The period of the orbit,tau=2.507 h